Session 2: Homework 1
Group 3
2021-09-04
1 Where Do People Drink The Most Beer, Wine And Spirits?
Back in 2014, fivethiryeight.com published an article on alchohol consumption in different countries. The data drinks is available as part of the fivethirtyeight package. Make sure you have installed the fivethirtyeight package before proceeding.
library(fivethirtyeight)
data(drinks)
# or download directly
alcohol_direct <- read_csv("https://raw.githubusercontent.com/fivethirtyeight/data/master/alcohol-consumption/drinks.csv")What are the variable types? Any missing values we should worry about?
glimpse(alcohol_direct
)## Rows: 193
## Columns: 5
## $ country <chr> "Afghanistan", "Albania", "Algeria", "And…
## $ beer_servings <dbl> 0, 89, 25, 245, 217, 102, 193, 21, 261, 2…
## $ spirit_servings <dbl> 0, 132, 0, 138, 57, 128, 25, 179, 72, 75,…
## $ wine_servings <dbl> 0, 54, 14, 312, 45, 45, 221, 11, 212, 191…
## $ total_litres_of_pure_alcohol <dbl> 0.0, 4.9, 0.7, 12.4, 5.9, 4.9, 8.3, 3.8, …skim(alcohol_direct)| Name | alcohol_direct |
| Number of rows | 193 |
| Number of columns | 5 |
| _______________________ | |
| Column type frequency: | |
| character | 1 |
| numeric | 4 |
| ________________________ | |
| Group variables | None |
Variable type: character
| skim_variable | n_missing | complete_rate | min | max | empty | n_unique | whitespace |
|---|---|---|---|---|---|---|---|
| country | 0 | 1 | 3 | 28 | 0 | 193 | 0 |
Variable type: numeric
| skim_variable | n_missing | complete_rate | mean | sd | p0 | p25 | p50 | p75 | p100 | hist |
|---|---|---|---|---|---|---|---|---|---|---|
| beer_servings | 0 | 1 | 106.16 | 101.14 | 0 | 20.0 | 76.0 | 188.0 | 376.0 | ▇▃▂▂▁ |
| spirit_servings | 0 | 1 | 80.99 | 88.28 | 0 | 4.0 | 56.0 | 128.0 | 438.0 | ▇▃▂▁▁ |
| wine_servings | 0 | 1 | 49.45 | 79.70 | 0 | 1.0 | 8.0 | 59.0 | 370.0 | ▇▁▁▁▁ |
| total_litres_of_pure_alcohol | 0 | 1 | 4.72 | 3.77 | 0 | 1.3 | 4.2 | 7.2 | 14.4 | ▇▃▅▃▁ |
Make a plot that shows the top 25 beer consuming countries
# YOUR CODE GOES HERE
p <- alcohol_direct %>%
slice_max(order_by = beer_servings, n = 25)
ggplot(p, aes(y=fct_reorder(country,beer_servings), x=beer_servings))+
geom_col()+
theme_bw( )+
labs(title = "", x="Beer Consumption", y="Country")
Make a plot that shows the top 25 wine consuming countries
p2 <- alcohol_direct %>%
slice_max(order_by = wine_servings, n = 25)
ggplot(p2, aes(y=fct_reorder(country,wine_servings), x=wine_servings))+
geom_col(fill="darkred")+
theme_bw( )+
labs(title = "", x="Wine Consumption", y="Country")
Finally, make a plot that shows the top 25 spirit consuming countries
# YOUR CODE GOES HERE
p3 <- alcohol_direct %>%
slice_max(order_by = spirit_servings, n = 25)
ggplot(p3, aes(y=fct_reorder(country,spirit_servings), x=spirit_servings))+
geom_col()+
theme_bw( )+
labs(title = "", x="Spirit Consumption", y="Country")
What can you infer from these plots? Don’t just explain what’s in the graph, but speculate or tell a short story (1-2 paragraphs max).
TYPE YOUR ANSWER AFTER (AND OUTSIDE!) THIS BLOCKQUOTE.
In general, there tends to be no overlap among countries with regards to the top spots in alcohol consumption, i.e. the top beer consuming countries are not the top wine consuming countries. It is surprising that Germany is only ranked #4 in the top beer consuming countries, after Namibia, Czech Republic and Gabon. Also, one can observe that the top wine consuming countries are almost exclusively European, which might be due to the reason that wine production is also among the highest in these countries and thus culturally rooted.
2 Analysis of movies- IMDB dataset
We will look at a subset sample of movies, taken from the Kaggle IMDB 5000 movie dataset
movies <- read_csv(here::here("data", "movies.csv"))
glimpse(movies)## Rows: 2,961
## Columns: 11
## $ title <chr> "Avatar", "Titanic", "Jurassic World", "The Avenge…
## $ genre <chr> "Action", "Drama", "Action", "Action", "Action", "…
## $ director <chr> "James Cameron", "James Cameron", "Colin Trevorrow…
## $ year <dbl> 2009, 1997, 2015, 2012, 2008, 1999, 1977, 2015, 20…
## $ duration <dbl> 178, 194, 124, 173, 152, 136, 125, 141, 164, 93, 1…
## $ gross <dbl> 7.61e+08, 6.59e+08, 6.52e+08, 6.23e+08, 5.33e+08, …
## $ budget <dbl> 2.37e+08, 2.00e+08, 1.50e+08, 2.20e+08, 1.85e+08, …
## $ cast_facebook_likes <dbl> 4834, 45223, 8458, 87697, 57802, 37723, 13485, 920…
## $ votes <dbl> 886204, 793059, 418214, 995415, 1676169, 534658, 9…
## $ reviews <dbl> 3777, 2843, 1934, 2425, 5312, 3917, 1752, 1752, 35…
## $ rating <dbl> 7.9, 7.7, 7.0, 8.1, 9.0, 6.5, 8.7, 7.5, 8.5, 7.2, …Besides the obvious variables of title, genre, director, year, and duration, the rest of the variables are as follows:
gross: The gross earnings in the US box office, not adjusted for inflationbudget: The movie’s budgetcast_facebook_likes: the number of facebook likes cast memebrs receivedvotes: the number of people who voted for (or rated) the movie in IMDBreviews: the number of reviews for that movierating: IMDB average rating
2.1 Use your data import, inspection, and cleaning skills to answer the following:
- Are there any missing values (NAs)? Are all entries distinct or are there duplicate entries?
skim(movies)| Name | movies |
| Number of rows | 2961 |
| Number of columns | 11 |
| _______________________ | |
| Column type frequency: | |
| character | 3 |
| numeric | 8 |
| ________________________ | |
| Group variables | None |
Variable type: character
| skim_variable | n_missing | complete_rate | min | max | empty | n_unique | whitespace |
|---|---|---|---|---|---|---|---|
| title | 0 | 1 | 1 | 83 | 0 | 2907 | 0 |
| genre | 0 | 1 | 5 | 11 | 0 | 17 | 0 |
| director | 0 | 1 | 3 | 32 | 0 | 1366 | 0 |
Variable type: numeric
| skim_variable | n_missing | complete_rate | mean | sd | p0 | p25 | p50 | p75 | p100 | hist |
|---|---|---|---|---|---|---|---|---|---|---|
| year | 0 | 1 | 2.00e+03 | 9.95e+00 | 1920.0 | 2.00e+03 | 2.00e+03 | 2.01e+03 | 2.02e+03 | ▁▁▁▂▇ |
| duration | 0 | 1 | 1.10e+02 | 2.22e+01 | 37.0 | 9.50e+01 | 1.06e+02 | 1.19e+02 | 3.30e+02 | ▃▇▁▁▁ |
| gross | 0 | 1 | 5.81e+07 | 7.25e+07 | 703.0 | 1.23e+07 | 3.47e+07 | 7.56e+07 | 7.61e+08 | ▇▁▁▁▁ |
| budget | 0 | 1 | 4.06e+07 | 4.37e+07 | 218.0 | 1.10e+07 | 2.60e+07 | 5.50e+07 | 3.00e+08 | ▇▂▁▁▁ |
| cast_facebook_likes | 0 | 1 | 1.24e+04 | 2.05e+04 | 0.0 | 2.24e+03 | 4.60e+03 | 1.69e+04 | 6.57e+05 | ▇▁▁▁▁ |
| votes | 0 | 1 | 1.09e+05 | 1.58e+05 | 5.0 | 1.99e+04 | 5.57e+04 | 1.33e+05 | 1.69e+06 | ▇▁▁▁▁ |
| reviews | 0 | 1 | 5.03e+02 | 4.94e+02 | 2.0 | 1.99e+02 | 3.64e+02 | 6.31e+02 | 5.31e+03 | ▇▁▁▁▁ |
| rating | 0 | 1 | 6.39e+00 | 1.05e+00 | 1.6 | 5.80e+00 | 6.50e+00 | 7.10e+00 | 9.30e+00 | ▁▁▆▇▁ |
There are no missing values, however, there are duplicate values for some variables. What is especially interesting, is that there are duplicate Titles for movies. This can be observed through looking at n_unique: Even though there are a total of 2961 records, there only seem to be 2907 unique movie titles.
- Produce a table with the count of movies by genre, ranked in descending order
count3<-movies%>%
group_by(genre)%>%
count(sort=TRUE)%>%
rename(count=n)
count3## # A tibble: 17 × 2
## # Groups: genre [17]
## genre count
## <chr> <int>
## 1 Comedy 848
## 2 Action 738
## 3 Drama 498
## 4 Adventure 288
## 5 Crime 202
## 6 Biography 135
## 7 Horror 131
## 8 Animation 35
## 9 Fantasy 28
## 10 Documentary 25
## 11 Mystery 16
## 12 Sci-Fi 7
## 13 Family 3
## 14 Musical 2
## 15 Romance 2
## 16 Western 2
## 17 Thriller 1- Produce a table with the average gross earning and budget (
grossandbudget) by genre. Calculate a variablereturn_on_budgetwhich shows how many $ did a movie make at the box office for each $ of its budget. Ranked genres by thisreturn_on_budgetin descending order
avgGenre <- movies%>%
group_by(genre)%>%
summarise(mean(gross), mean(budget))%>%
rename(
"Average_Gross" = "mean(gross)",
"Average_Budget" = "mean(budget)"
)%>%
mutate(return_on_budget = Average_Gross/Average_Budget)%>%
arrange(desc(return_on_budget))
#format(avgGenre$Average_Gross, big.mark=".", scientific=FALSE)
options(scipen = 50)
avgGenre## # A tibble: 17 × 4
## genre Average_Gross Average_Budget return_on_budget
## <chr> <dbl> <dbl> <dbl>
## 1 Musical 92084000 3189500 28.9
## 2 Family 149160478. 14833333. 10.1
## 3 Western 20821884 3465000 6.01
## 4 Documentary 17353973. 5887852. 2.95
## 5 Horror 37713738. 13504916. 2.79
## 6 Fantasy 42408841. 17582143. 2.41
## 7 Comedy 42630552. 24446319. 1.74
## 8 Mystery 67533021. 39218750 1.72
## 9 Animation 98433792. 61701429. 1.60
## 10 Biography 45201805. 28543696. 1.58
## 11 Adventure 95794257. 66290069. 1.45
## 12 Drama 37465371. 26242933. 1.43
## 13 Crime 37502397. 26596169. 1.41
## 14 Romance 31264848. 25107500 1.25
## 15 Action 86583860. 71354888. 1.21
## 16 Sci-Fi 29788371. 27607143. 1.08
## 17 Thriller 2468 300000 0.00823- Produce a table that shows the top 15 directors who have created the highest gross revenue in the box office. Don’t just show the total gross amount, but also the mean, median, and standard deviation per director.
topDirectors <- movies%>%
group_by(director)%>%
summarise(total_gross = sum(gross), avg_gross = mean(gross), median_gross = median(gross), sd_gross = sd(gross))%>%
slice_max(order_by = total_gross, n = 15)
topDirectors## # A tibble: 15 × 5
## director total_gross avg_gross median_gross sd_gross
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 Steven Spielberg 4014061704 174524422. 164435221 101421051.
## 2 Michael Bay 2231242537 171634041. 138396624 127161579.
## 3 Tim Burton 2071275480 129454718. 76519172 108726924.
## 4 Sam Raimi 2014600898 201460090. 234903076 162126632.
## 5 James Cameron 1909725910 318287652. 175562880. 309171337.
## 6 Christopher Nolan 1813227576 226653447 196667606. 187224133.
## 7 George Lucas 1741418480 348283696 380262555 146193880.
## 8 Robert Zemeckis 1619309108 124562239. 100853835 91300279.
## 9 Clint Eastwood 1378321100 72543216. 46700000 75487408.
## 10 Francis Lawrence 1358501971 271700394. 281666058 135437020.
## 11 Ron Howard 1335988092 111332341 101587923 81933761.
## 12 Gore Verbinski 1329600995 189942999. 123207194 154473822.
## 13 Andrew Adamson 1137446920 284361730 279680930. 120895765.
## 14 Shawn Levy 1129750988 102704635. 85463309 65484773.
## 15 Ridley Scott 1128857598 80632686. 47775715 68812285.- Finally, ratings. Produce a table that describes how ratings are distributed by genre. We don’t want just the mean, but also, min, max, median, SD and some kind of a histogram or density graph that visually shows how ratings are distributed.
ratings <- movies%>%
group_by(genre)%>%
summarise(avg_rating = mean(rating), min_rating = min(rating), max_rating = max(rating), median_rating = median(rating), sd_rating = sd(rating))%>%
arrange(desc(avg_rating))
ratings## # A tibble: 17 × 6
## genre avg_rating min_rating max_rating median_rating sd_rating
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 Biography 7.11 4.5 8.9 7.2 0.760
## 2 Crime 6.92 4.8 9.3 6.9 0.849
## 3 Mystery 6.86 4.6 8.5 6.9 0.882
## 4 Musical 6.75 6.3 7.2 6.75 0.636
## 5 Drama 6.73 2.1 8.8 6.8 0.917
## 6 Documentary 6.66 1.6 8.5 7.4 1.77
## 7 Sci-Fi 6.66 5 8.2 6.4 1.09
## 8 Animation 6.65 4.5 8 6.9 0.968
## 9 Romance 6.65 6.2 7.1 6.65 0.636
## 10 Adventure 6.51 2.3 8.6 6.6 1.09
## 11 Family 6.5 5.7 7.9 5.9 1.22
## 12 Action 6.23 2.1 9 6.3 1.03
## 13 Fantasy 6.15 4.3 7.9 6.45 0.959
## 14 Comedy 6.11 1.9 8.8 6.2 1.02
## 15 Horror 5.83 3.6 8.5 5.9 1.01
## 16 Western 5.7 4.1 7.3 5.7 2.26
## 17 Thriller 4.8 4.8 4.8 4.8 NAoverall<-ggplot(movies, aes(x=rating))+
geom_histogram(color="black", fill="white")+
geom_vline(aes(xintercept=mean(rating), color = "red"))+
labs(title = "overall rating distribution")
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rating_by_genre<-ggplot(movies, aes(x=rating))+
geom_histogram(color="black", fill="white")+
facet_wrap(vars(genre))
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2.2 Use ggplot to answer the following
- Examine the relationship between
grossandcast_facebook_likes. Produce a scatterplot and write one sentence discussing whether the number of facebook likes that the cast has received is likely to be a good predictor of how much money a movie will make at the box office. What variable are you going to map to the Y- and X- axes?
all_scatter_likes <- ggplot(movies, aes(x=cast_facebook_likes, y=gross))+
geom_point(aes(color=genre))+
geom_smooth(method = "lm")+
theme_bw()
all_scatter_likes
#we left out the visualization of outlying values in order to make it easier to read
adjusted_scatter_likes <- ggplot(movies,aes(x=cast_facebook_likes, y=gross))+
# add = "reg.line" # Add regressin line
# add.params = list(color = "blue", fill = "lightgray"), # Customize reg. line
# conf.int = TRUE # Add confidence interval
#
geom_point(aes(color=genre))+
geom_smooth(method = "lm")+
xlim(0,150000)+
ylim(0,500000000)+
theme_bw()
adjusted_scatter_likes
From the trendline and correlation coefficient of 0.28 one can observe that there is a weak positive correlation between the two variables indicating that the casts’ Facebook popularity may help predict the gross revenue of a movie to a slight extent.
- Examine the relationship between
grossandbudget. Produce a scatterplot and write one sentence discussing whether budget is likely to be a good predictor of how much money a movie will make at the box office.
all_scatter_budget <- ggplot(movies, aes(x=budget, y=gross))+
geom_point(aes(color=genre))+
geom_smooth(method = "lm")+
theme_bw()
all_scatter_budget
Compared to the casts’ facebook popularity, budget seems to be a stronger predictor of a movie’s gross revenue as the correlation coefficient is closer to 1.
- Examine the relationship between
grossandrating. Produce a scatterplot, faceted bygenreand discuss whether IMDB ratings are likely to be a good predictor of how much money a movie will make at the box office. Is there anything strange in this dataset?
all_scatter_rating <- ggplot(movies, aes(x=rating, y=gross))+
geom_point(aes(color=genre), size=0.5)+
ylim(0,750000000)+
geom_smooth(method = "lm")+
facet_wrap(vars(genre))+
theme_bw()
all_scatter_rating
Overall, one can observe a positive correlation between a movie’s rating and its gross revenue, however, as with casts’ facebook likes, the correlation is weak. Due to the small sample size of some genres (e.g. Musical, Western and Sci-Fi), not all trendlines are representative.
3 Returns of financial stocks
You may find useful the material on finance data sources.
We will use the tidyquant package to download historical data of stock prices, calculate returns, and examine the distribution of returns.
We must first identify which stocks we want to download data for, and for this we must know their ticker symbol; Apple is known as AAPL, Microsoft as MSFT, McDonald’s as MCD, etc. The file nyse.csv contains 508 stocks listed on the NYSE, their ticker symbol, name, the IPO (Initial Public Offering) year, and the sector and industry the company is in.
nyse <- read_csv(here::here("data","nyse.csv"))Based on this dataset, create a table and a bar plot that shows the number of companies per sector, in descending order
# YOUR CODE GOES HERE
sector<-nyse%>%
group_by(sector)%>%
summarise(number = count(sector))%>%
arrange(desc(number))
sector## # A tibble: 12 × 2
## sector number
## <chr> <int>
## 1 Finance 97
## 2 Consumer Services 79
## 3 Public Utilities 60
## 4 Capital Goods 45
## 5 Health Care 45
## 6 Energy 42
## 7 Technology 40
## 8 Basic Industries 39
## 9 Consumer Non-Durables 31
## 10 Miscellaneous 12
## 11 Transportation 10
## 12 Consumer Durables 8barplot<- ggplot(sector, aes(y=fct_reorder(sector, number), x=number))+
geom_col(width = 0.5)+
labs(title = "companies per sector", y = "sector", x= "number of companies")+
theme_bw()
barplot
Next, let’s choose some stocks and their ticker symbols and download some data. You MUST choose 6 different stocks from the ones listed below; You should, however, add SPY which is the SP500 ETF (Exchange Traded Fund).
# Notice the cache=TRUE argument in the chunk options. Because getting data is time consuming,
# cache=TRUE means that once it downloads data, the chunk will not run again next time you knit your Rmd
myStocks <- c("GREK","GME","AMC","AAPL","COW","TSLA","SPY" ) %>%
tq_get(get = "stock.prices",
from = "2011-01-01",
to = "2021-08-31") %>%
group_by(symbol)
glimpse(myStocks) # examine the structure of the resulting data frame## Rows: 16,018
## Columns: 8
## Groups: symbol [7]
## $ symbol <chr> "GREK", "GREK", "GREK", "GREK", "GREK", "GREK", "GREK", "GREK…
## $ date <date> 2011-12-08, 2011-12-09, 2011-12-12, 2011-12-13, 2011-12-14, …
## $ open <dbl> 44.8, 44.1, 44.4, 44.8, 41.9, 41.8, 42.0, 40.9, 42.3, 42.1, 4…
## $ high <dbl> 44.9, 44.7, 44.4, 44.8, 41.9, 42.0, 42.2, 43.4, 42.3, 42.1, 4…
## $ low <dbl> 44.2, 44.1, 42.5, 42.0, 40.8, 41.8, 39.9, 40.4, 41.4, 42.1, 4…
## $ close <dbl> 44.3, 44.4, 42.7, 42.3, 41.5, 42.0, 41.2, 41.0, 41.4, 42.1, 4…
## $ volume <dbl> 1167, 667, 2700, 2633, 200, 1267, 3667, 1467, 1100, 167, 700,…
## $ adjusted <dbl> 38.3, 38.4, 36.9, 36.6, 36.0, 36.3, 35.7, 35.5, 35.8, 36.5, 3…Financial performance analysis depend on returns; If I buy a stock today for 100 and I sell it tomorrow for 101.75, my one-day return, assuming no transaction costs, is 1.75%. So given the adjusted closing prices, our first step is to calculate daily and monthly returns.
#calculate daily returns
myStocks_returns_daily <- myStocks %>%
tq_transmute(select = adjusted,
mutate_fun = periodReturn,
period = "daily",
type = "log",
col_rename = "daily_returns",
cols = c(nested.col))
#calculate monthly returns
myStocks_returns_monthly <- myStocks %>%
tq_transmute(select = adjusted,
mutate_fun = periodReturn,
period = "monthly",
type = "arithmetic",
col_rename = "monthly_returns",
cols = c(nested.col))
#calculate yearly returns
myStocks_returns_annual <- myStocks %>%
group_by(symbol) %>%
tq_transmute(select = adjusted,
mutate_fun = periodReturn,
period = "yearly",
type = "arithmetic",
col_rename = "yearly_returns",
cols = c(nested.col))Create a table where you summarise monthly returns for each of the stocks and SPY; min, max, median, mean, SD.
# YOUR CODE GOES HERE
summary_monthly<-myStocks_returns_monthly%>%
group_by(symbol)%>%
summarise(min_return = min(monthly_returns), max_return = max(monthly_returns), median_return = median(monthly_returns), avg_return = mean(monthly_returns), sd_return = sd(monthly_returns))
summary_monthly## # A tibble: 7 × 6
## symbol min_return max_return median_return avg_return sd_return
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 AAPL -0.181 0.217 0.0257 0.0245 0.0785
## 2 AMC -0.504 5.25 0.00679 0.0798 0.611
## 3 COW -0.134 0.0762 0.00264 -0.00524 0.0485
## 4 GME -0.687 16.3 -0.0118 0.142 1.45
## 5 GREK -0.332 0.369 0.00609 0.00365 0.110
## 6 SPY -0.125 0.127 0.0174 0.0123 0.0381
## 7 TSLA -0.224 0.811 0.0148 0.0523 0.176Plot a density plot, using geom_density(), for each of the stocks
# YOUR CODE GOES HERE
density<-ggplot(myStocks_returns_monthly, aes(x=monthly_returns))+
geom_density()+
facet_wrap(vars(symbol), scales="free")
density
What can you infer from this plot? Which stock is the riskiest? The least risky?
TYPE YOUR ANSWER AFTER (AND OUTSIDE!) THIS BLOCKQUOTE.
From the density plots, one can observe that volatility in monthly return varies widely across the stocks/ETF. For instance, the returns of COW seem to be more evenly distributed than GME, indicating a lower risk when investing into them. In our eyes, GME seems to be the riskiest stock, while an investment into SPY carries the least risk. The latter is understandable as the instrument holds shares of all S&P500 companies and is thus more diversified.
Looking at the x axis, one can also observe a large difference in the range of monthly returns across the instruments. Interestingly, most of the distribution lies around the 0 mark for all displayed tickers.
Finally, make a plot that shows the expected monthly return (mean) of a stock on the Y axis and the risk (standard deviation) in the X-axis. Please use ggrepel::geom_text_repel() to label each stock
# YOUR CODE GOES HERE
scatter_return_volatility <- ggplot(summary_monthly, aes(x=sd_return, y=avg_return, label=symbol))+
geom_point()+
geom_text_repel()
scatter_return_volatility
What can you infer from this plot? Are there any stocks which, while being riskier, do not have a higher expected return?
TYPE YOUR ANSWER AFTER (AND OUTSIDE!) THIS BLOCKQUOTE.
Generally, the market seems to reward higher risk with higher return. However, there are minor exceptions in our sample: GREK, indexing the Greek market, has a higher SD than SPY, yet its average monthly return is lower.
4 On your own: IBM HR Analytics
For this task, you will analyse a data set on Human Resoruce Analytics. The IBM HR Analytics Employee Attrition & Performance data set is a fictional data set created by IBM data scientists. Among other things, the data set includes employees’ income, their distance from work, their position in the company, their level of education, etc. A full description can be found on the website.
First let us load the data
hr_dataset <- read_csv(here::here("data", "datasets_1067_1925_WA_Fn-UseC_-HR-Employee-Attrition.csv"))
glimpse(hr_dataset)## Rows: 1,470
## Columns: 35
## $ Age <dbl> 41, 49, 37, 33, 27, 32, 59, 30, 38, 36, 35, 2…
## $ Attrition <chr> "Yes", "No", "Yes", "No", "No", "No", "No", "…
## $ BusinessTravel <chr> "Travel_Rarely", "Travel_Frequently", "Travel…
## $ DailyRate <dbl> 1102, 279, 1373, 1392, 591, 1005, 1324, 1358,…
## $ Department <chr> "Sales", "Research & Development", "Research …
## $ DistanceFromHome <dbl> 1, 8, 2, 3, 2, 2, 3, 24, 23, 27, 16, 15, 26, …
## $ Education <dbl> 2, 1, 2, 4, 1, 2, 3, 1, 3, 3, 3, 2, 1, 2, 3, …
## $ EducationField <chr> "Life Sciences", "Life Sciences", "Other", "L…
## $ EmployeeCount <dbl> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, …
## $ EmployeeNumber <dbl> 1, 2, 4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16,…
## $ EnvironmentSatisfaction <dbl> 2, 3, 4, 4, 1, 4, 3, 4, 4, 3, 1, 4, 1, 2, 3, …
## $ Gender <chr> "Female", "Male", "Male", "Female", "Male", "…
## $ HourlyRate <dbl> 94, 61, 92, 56, 40, 79, 81, 67, 44, 94, 84, 4…
## $ JobInvolvement <dbl> 3, 2, 2, 3, 3, 3, 4, 3, 2, 3, 4, 2, 3, 3, 2, …
## $ JobLevel <dbl> 2, 2, 1, 1, 1, 1, 1, 1, 3, 2, 1, 2, 1, 1, 1, …
## $ JobRole <chr> "Sales Executive", "Research Scientist", "Lab…
## $ JobSatisfaction <dbl> 4, 2, 3, 3, 2, 4, 1, 3, 3, 3, 2, 3, 3, 4, 3, …
## $ MaritalStatus <chr> "Single", "Married", "Single", "Married", "Ma…
## $ MonthlyIncome <dbl> 5993, 5130, 2090, 2909, 3468, 3068, 2670, 269…
## $ MonthlyRate <dbl> 19479, 24907, 2396, 23159, 16632, 11864, 9964…
## $ NumCompaniesWorked <dbl> 8, 1, 6, 1, 9, 0, 4, 1, 0, 6, 0, 0, 1, 0, 5, …
## $ Over18 <chr> "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", …
## $ OverTime <chr> "Yes", "No", "Yes", "Yes", "No", "No", "Yes",…
## $ PercentSalaryHike <dbl> 11, 23, 15, 11, 12, 13, 20, 22, 21, 13, 13, 1…
## $ PerformanceRating <dbl> 3, 4, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3, 3, 3, 3, …
## $ RelationshipSatisfaction <dbl> 1, 4, 2, 3, 4, 3, 1, 2, 2, 2, 3, 4, 4, 3, 2, …
## $ StandardHours <dbl> 80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 8…
## $ StockOptionLevel <dbl> 0, 1, 0, 0, 1, 0, 3, 1, 0, 2, 1, 0, 1, 1, 0, …
## $ TotalWorkingYears <dbl> 8, 10, 7, 8, 6, 8, 12, 1, 10, 17, 6, 10, 5, 3…
## $ TrainingTimesLastYear <dbl> 0, 3, 3, 3, 3, 2, 3, 2, 2, 3, 5, 3, 1, 2, 4, …
## $ WorkLifeBalance <dbl> 1, 3, 3, 3, 3, 2, 2, 3, 3, 2, 3, 3, 2, 3, 3, …
## $ YearsAtCompany <dbl> 6, 10, 0, 8, 2, 7, 1, 1, 9, 7, 5, 9, 5, 2, 4,…
## $ YearsInCurrentRole <dbl> 4, 7, 0, 7, 2, 7, 0, 0, 7, 7, 4, 5, 2, 2, 2, …
## $ YearsSinceLastPromotion <dbl> 0, 1, 0, 3, 2, 3, 0, 0, 1, 7, 0, 0, 4, 1, 0, …
## $ YearsWithCurrManager <dbl> 5, 7, 0, 0, 2, 6, 0, 0, 8, 7, 3, 8, 3, 2, 3, …I am going to clean the data set, as variable names are in capital letters, some variables are not really necessary, and some variables, e.g., education are given as a number rather than a more useful description
hr_cleaned <- hr_dataset %>%
clean_names() %>%
mutate(
education = case_when(
education == 1 ~ "Below College",
education == 2 ~ "College",
education == 3 ~ "Bachelor",
education == 4 ~ "Master",
education == 5 ~ "Doctor"
),
environment_satisfaction = case_when(
environment_satisfaction == 1 ~ "Low",
environment_satisfaction == 2 ~ "Medium",
environment_satisfaction == 3 ~ "High",
environment_satisfaction == 4 ~ "Very High"
),
job_satisfaction = case_when(
job_satisfaction == 1 ~ "Low",
job_satisfaction == 2 ~ "Medium",
job_satisfaction == 3 ~ "High",
job_satisfaction == 4 ~ "Very High"
),
performance_rating = case_when(
performance_rating == 1 ~ "Low",
performance_rating == 2 ~ "Good",
performance_rating == 3 ~ "Excellent",
performance_rating == 4 ~ "Outstanding"
),
work_life_balance = case_when(
work_life_balance == 1 ~ "Bad",
work_life_balance == 2 ~ "Good",
work_life_balance == 3 ~ "Better",
work_life_balance == 4 ~ "Best"
)
) %>%
select(age, attrition, daily_rate, department,
distance_from_home, education,
gender, job_role,environment_satisfaction,
job_satisfaction, marital_status,
monthly_income, num_companies_worked, percent_salary_hike,
performance_rating, total_working_years,
work_life_balance, years_at_company,
years_since_last_promotion)Produce a one-page summary describing this dataset. Here is a non-exhaustive list of questions:
- How often do people leave the company (
attrition)
numberz<-hr_cleaned%>%
group_by(attrition)%>%
summarize(part = count(attrition))%>%
mutate(perc = part/sum(part))
NULL## NULLnumberz## # A tibble: 2 × 3
## attrition part perc
## <chr> <int> <dbl>
## 1 No 1233 0.839
## 2 Yes 237 0.161Looking at the sheer counts of YES and NO entries for the attrition coloumn, around 237 of all employees have left the company, while 1233 have been staying employed. This corresponds to around 16% of all recorded employees.
- How are
age,years_at_company,monthly_incomeandyears_since_last_promotiondistributed? can you roughly guess which of these variables is closer to Normal just by looking at summary statistics?
n<-length(hr_cleaned$age)
pic_data<-data.frame(cbind(c(hr_cleaned$age,hr_cleaned$monthly_income,
hr_cleaned$years_since_last_promotion),
c(rep("age",n),rep("monthly_income",n),rep("years_since_last_promotion",n))))
colnames(pic_data)<-c("Num","Attribute")
pic_data$Num<-as.integer(pic_data$Num)
colnames(pic_data)<-c("Num","Attribute")
distribution<-ggplot(pic_data,aes(x=Num))+
geom_histogram()+
facet_wrap(vars(Attribute),scale="free")+
labs(x="",y="Frequency")+
theme_bw()
distribution 
- How are
job_satisfactionandwork_life_balancedistributed? Don’t just report counts, but express categories as % of total
hr_satis <- hr_cleaned %>%
group_by(job_satisfaction) %>%
count(sort=TRUE) %>%
mutate(percent=n/1470)
hr_satis ## # A tibble: 4 × 3
## # Groups: job_satisfaction [4]
## job_satisfaction n percent
## <chr> <int> <dbl>
## 1 Very High 459 0.312
## 2 High 442 0.301
## 3 Low 289 0.197
## 4 Medium 280 0.190work_life_satis <- hr_cleaned %>%
group_by(work_life_balance) %>%
count(sort=TRUE) %>%
mutate(percent=n/1470)
work_life_satis ## # A tibble: 4 × 3
## # Groups: work_life_balance [4]
## work_life_balance n percent
## <chr> <int> <dbl>
## 1 Better 893 0.607
## 2 Good 344 0.234
## 3 Best 153 0.104
## 4 Bad 80 0.0544- Is there any relationship between monthly income and education? Monthly income and gender?
plot1 <- ggplot(hr_cleaned, aes(y=monthly_income, x = education))+
geom_boxplot()
plot1 
plot2 <-ggplot(hr_cleaned, aes(y=monthly_income, x = gender))+
geom_boxplot()
plot2 
- Plot a boxplot of income vs job role. Make sure the highest-paid job roles appear first
income_vs_job <- ggplot(hr_cleaned, aes(x=reorder(job_role, -monthly_income, na.rm = TRUE), y=monthly_income)) +
geom_boxplot()+
labs(y="Monthly Income", x = "Job", title = "Income Distribution By Job")+
theme(axis.text.x = element_text(angle = 90, vjust = 0.5, hjust=1))
income_vs_job
- Calculate and plot a bar chart of the mean (or median?) income by education level.
ggplo <- ggplot(hr_cleaned, aes(x=edu))- Plot the distribution of income by education level. Use a facet_wrap and a theme from
ggthemes
plot7 <- hr_cleaned %>%
group_by(education) %>%
ggplot(aes(x=monthly_income))+
geom_histogram()+
ggthemes::theme_wsj()
plot7 
- Plot income vs age, faceted by
job_role
plot8 <- ggplot(hr_cleaned, aes(x=age, y=monthly_income))+
geom_point(size=0.1)+
geom_smooth(method = "lm")+
facet_wrap(vars(job_role), scales = "free" )
plot8 
# Challenge 1: Replicating a chart
The purpose of this exercise is to reproduce a plot using your dplyr and ggplot2 skills. Read the article The Racial Factor: There’s 77 Counties Which Are Deep Blue But Also Low-Vaxx. Guess What They Have In Common? and have a look at the attached figure.
You dont have to worry about the blue-red backgound and don’t worry about replicating it exactly, try and see how far you can get. You’re encouraged to work together if you want to and exchange tips/tricks you figured out– and even though the figure in the original article is from early July 2021, you can use the most recent data.
Some hints to get you started:
- To get vaccination by county, we will use data from the CDC
- You need to get County Presidential Election Returns 2000-2020
- Finally, you also need an estimate of the population of each county
skim(population)| Name | population |
| Number of rows | 3273 |
| Number of columns | 2 |
| _______________________ | |
| Column type frequency: | |
| character | 1 |
| numeric | 1 |
| ________________________ | |
| Group variables | None |
Variable type: character
| skim_variable | n_missing | complete_rate | min | max | empty | n_unique | whitespace |
|---|---|---|---|---|---|---|---|
| fips | 0 | 1 | 5 | 5 | 0 | 3273 | 0 |
Variable type: numeric
| skim_variable | n_missing | complete_rate | mean | sd | p0 | p25 | p50 | p75 | p100 | hist |
|---|---|---|---|---|---|---|---|---|---|---|
| pop_estimate_2019 | 0 | 1 | 302813 | 5867752 | 86 | 11317 | 26687 | 71783 | 328239523 | ▇▁▁▁▁ |
#as there are multiple dates for vacc statuses, and we are only interested in the most recent one, we can do some cleaning. Data from Sep. 3rd, 2021, seems to be most recent.
vaccinations_recent <- vaccinations %>%
filter(date=="09/03/2021")
#we can also include an additional coloumn to the election dataset, indicating the number of votes for one candidate as a percentage of whole
elections_perc <- election2020_results%>%
mutate(perc_votes = candidatevotes/totalvotes)#We decide to just join coloumn series_complete_pop_pct from the vacc dataset, as this should suffice to recreate the graph. We also rename the coloumns to make their names a little more easy on the tongue.
joined_frame <- inner_join(population, vaccinations_recent[,c("fips", "series_complete_pop_pct") ], by = "fips")%>%
rename("vacc_perc" = "series_complete_pop_pct", "pop" = "pop_estimate_2019")
#Similarly, we join parts of the cleaned election data, and filter out records other than votes for Trump, as they are not needed for the graph. In addtion, we sum up perc_votes because there is sometimes data for different ways of voting, i.e. early vote, election day, provisional.
joined_frame <- inner_join(joined_frame, elections_perc[,c("fips", "candidate", "perc_votes")])%>%
filter(candidate=="DONALD J TRUMP")%>%
group_by(fips, pop, vacc_perc, candidate, )%>%
summarise(sum(perc_votes))%>%
rename("perc_votes_total" = "sum(perc_votes)")
head(joined_frame) #we can have a look at the dataset. It looks like we have all the data we need for visualization## # A tibble: 6 × 5
## # Groups: fips, pop, vacc_perc [6]
## fips pop vacc_perc candidate perc_votes_total
## <chr> <dbl> <dbl> <chr> <dbl>
## 1 01001 55869 30.5 DONALD J TRUMP 0.714
## 2 01003 223234 37.9 DONALD J TRUMP 0.762
## 3 01005 24686 31.3 DONALD J TRUMP 0.535
## 4 01007 22394 26.6 DONALD J TRUMP 0.784
## 5 01009 57826 22.6 DONALD J TRUMP 0.896
## 6 01011 10101 38.4 DONALD J TRUMP 0.248plot_counties <- ggplot(joined_frame, aes(x=perc_votes_total, vacc_perc))+
geom_rect(aes(xmin = 0, xmax = 0.45, ymin = 0, ymax = 100), fill = "#A6A8FB")+
geom_rect(aes(xmin = 0.45, xmax = 0.55, ymin = 0, ymax = 100), fill = "#D2A6D2")+
geom_rect(aes(xmin = 0.55, xmax = 1, ymin = 0, ymax = 100), fill = "#FDA7A7")+
geom_point(aes(size=pop), alpha = 1/3)+
geom_smooth(method = "lm")+
theme_bw()
plot_counties
#It seems like there are a number of outliers in the dataset. 276 counties have reported a full vaccination count of 0 for the Sep. 3rd collection. As you can see in the output, Bexar county, home to a population of 2003554 reported a vacc rate of 0%.
filter(joined_frame, fips==48029)## # A tibble: 1 × 5
## # Groups: fips, pop, vacc_perc [1]
## fips pop vacc_perc candidate perc_votes_total
## <chr> <dbl> <dbl> <chr> <dbl>
## 1 48029 2003554 0 DONALD J TRUMP 0.400#It might be appropriate, to exclude the records with vacc_perc = 0 from out dataset, as it seems very hard to believe that among 2003554 people, not a single one is fully vaccinated. An error like this potentially has a large impact on the regression line. We therefore create an adjusted DF.
adj_joined_fram <- joined_frame%>%
filter(vacc_perc!=0)
adj_plot_counties <- ggplot(adj_joined_fram, aes(x = perc_votes_total, y = vacc_perc))+
geom_rect(aes(xmin = 0, xmax = 0.45, ymin = 0, ymax = 100), fill = "#A6A8FB")+
geom_rect(aes(xmin = 0.45, xmax = 0.55, ymin = 0, ymax = 100), fill = "#D2A6D2")+
geom_rect(aes(xmin = 0.55, xmax = 1, ymin = 0, ymax = 100), fill = "#FDA7A7")+
geom_point(aes(size = pop), alpha = 1/3)+
geom_smooth(method = "lm")+
theme_bw()
adj_plot_counties
5 Challenge 2: Opinion polls for the 2021 German elections
The Guardian newspaper has an election poll tracker for the upcoming German election. The list of the opinion polls since Jan 2021 can be found at Wikipedia and your task is to reproduce the graph similar to the one produced by the Guardian.
The following code will scrape the wikipedia page and import the table in a dataframe.
url <- "https://en.wikipedia.org/wiki/Opinion_polling_for_the_2021_German_federal_election"
# https://www.economist.com/graphic-detail/who-will-succeed-angela-merkel
# https://www.theguardian.com/world/2021/jun/21/german-election-poll-tracker-who-will-be-the-next-chancellor
# get tables that exist on wikipedia page
tables <- url %>%
read_html() %>%
html_nodes(css="table")
# parse HTML tables into a dataframe called polls
# Use purr::map() to create a list of all tables in URL
polls <- map(tables, . %>%
html_table(fill=TRUE)%>%
janitor::clean_names())
# list of opinion polls
german_election_polls <- polls[[1]] %>% # the first table on the page contains the list of all opinions polls
slice(2:(n()-1)) %>% # drop the first row, as it contains again the variable names and last row that contains 2017 results
mutate(
# polls are shown to run from-to, e.g. 9-13 Aug 2021. We keep the last date, 13 Aug here, as the poll date
# and we extract it by picking the last 11 characters from that field
end_date = str_sub(fieldwork_date, -11),
# end_date is still a string, so we convert it into a date object using lubridate::dmy()
end_date = dmy(end_date),
# we also get the month and week number from the date, if we want to do analysis by month- week, etc.
month = month(end_date),
week = isoweek(end_date)
)plot_election_polls <- ggplot(german_election_polls, aes(x = end_date))+
geom_point(aes(y = union), color = "black", alpha = 0.3)+
geom_line(aes(y=rollmean(union, 14, na.pad=TRUE)), color = "black")+ #we add a line representing the rolling average of values for the past 14 days. We also adjust the padding for enhancing visualization.
geom_point(aes(y = spd), color = "red", alpha = 0.3)+
geom_line(aes(y=rollmean(spd, 14, na.pad=TRUE)), color = "red")+
geom_point(aes(y = grune), color = "darkgreen", alpha = 0.3)+
geom_line(aes(y=rollmean(grune, 14, na.pad=TRUE)), color = "darkgreen")+
geom_point(aes(y = af_d), color = "blue", alpha = 0.3)+
geom_line(aes(y=rollmean(af_d, 14, na.pad=TRUE)), color = "blue")+
geom_point(aes(y = fdp), color = "#F4E332", alpha = 0.3)+
geom_line(aes(y=rollmean(fdp, 14, na.pad=TRUE)), color = "#F4E332")+
geom_point(aes(y=linke), color = "darkred", alpha = 0.3)+
geom_line(aes(y=rollmean(linke, 14, na.pad = TRUE)), color = "darkred")+
labs(x="2021 Poll Month", y="Voting Percentage", title = "German Elections", subtitle = "Change in estimated voting during 2021")
theme_bw()## List of 93
## $ line :List of 6
## ..$ colour : chr "black"
## ..$ size : num 0.5
## ..$ linetype : num 1
## ..$ lineend : chr "butt"
## ..$ arrow : logi FALSE
## ..$ inherit.blank: logi TRUE
## ..- attr(*, "class")= chr [1:2] "element_line" "element"
## $ rect :List of 5
## ..$ fill : chr "white"
## ..$ colour : chr "black"
## ..$ size : num 0.5
## ..$ linetype : num 1
## ..$ inherit.blank: logi TRUE
## ..- attr(*, "class")= chr [1:2] "element_rect" "element"
## $ text :List of 11
## ..$ family : chr ""
## ..$ face : chr "plain"
## ..$ colour : chr "black"
## ..$ size : num 11
## ..$ hjust : num 0.5
## ..$ vjust : num 0.5
## ..$ angle : num 0
## ..$ lineheight : num 0.9
## ..$ margin : 'margin' num [1:4] 0points 0points 0points 0points
## .. ..- attr(*, "unit")= int 8
## ..$ debug : logi FALSE
## ..$ inherit.blank: logi TRUE
## ..- attr(*, "class")= chr [1:2] "element_text" "element"
## $ title : NULL
## $ aspect.ratio : NULL
## $ axis.title : NULL
## $ axis.title.x :List of 11
## ..$ family : NULL
## ..$ face : NULL
## ..$ colour : NULL
## ..$ size : NULL
## ..$ hjust : NULL
## ..$ vjust : num 1
## ..$ angle : NULL
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6 Deliverables
There is a lot of explanatory text, comments, etc. You do not need these, so delete them and produce a stand-alone document that you could share with someone. Knit the edited and completed R Markdown file as an HTML document (use the “Knit” button at the top of the script editor window) and upload it to Canvas.
7 Details
- Who did you collaborate with: TYPE NAMES HERE
- Approximately how much time did you spend on this problem set: ANSWER HERE
- What, if anything, gave you the most trouble: ANSWER HERE
Please seek out help when you need it, and remember the 15-minute rule. You know enough R (and have enough examples of code from class and your readings) to be able to do this. If you get stuck, ask for help from others, post a question on Slack– and remember that I am here to help too!
As a true test to yourself, do you understand the code you submitted and are you able to explain it to someone else?
8 Rubric
Check minus (1/5): Displays minimal effort. Doesn’t complete all components. Code is poorly written and not documented. Uses the same type of plot for each graph, or doesn’t use plots appropriate for the variables being analyzed.
Check (3/5): Solid effort. Hits all the elements. No clear mistakes. Easy to follow (both the code and the output).
Check plus (5/5): Finished all components of the assignment correctly and addressed both challenges. Code is well-documented (both self-documented and with additional comments as necessary). Used tidyverse, instead of base R. Graphs and tables are properly labelled. Analysis is clear and easy to follow, either because graphs are labeled clearly or you’ve written additional text to describe how you interpret the output.